Question: The tangent line to the graph of function $g$ at the point $(-6,-2)$ passes through the point $(0,2)$. Find $g'(-6)$. $g'(-6)=$
Solution: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $g'(-6)$ gives the slope of the tangent line to the graph of $g$ where $x=-6$, which is the point $(-6,-2)$. We know this line passes through $(-6,-2)$, and we are also given that it passes through $(0,2)$. This should be enough to find the slope of that line. $\begin{aligned} \text{Slope}&=\dfrac{\text{Change in }y}{\text{Change in }x} \\\\ &=\dfrac{2-(-2)}{0-(-6)} \\\\ &=\dfrac{4}{6} \\\\ &=\dfrac23 \end{aligned}$ In conclusion, $g'(-6)=\dfrac{2}{3}$.